# Derive a general expression for the phasor response I_L and V_o

#### cursed

Joined Oct 3, 2010
2
The circuit in the figure below is operating in the sinusoidal steady state with $$V_{s}(t)=V_{A} cos(\omega t)$$. Derive a general expression for the phasor response $$I_{L}$$ and the voltage $$V_{o}$$.

Relevant equations:

$$I= \frac{V_{A}}{Z_{EQ}}$$, where Z is the impedance and V is the volage.

My attempt:

I converted everything to the phasor ("frequency") domain. So the value of the inductor is now $$Z=j \omega L$$. The resistors do not change.

Now,

$$Z_{EQ} = R + (j \omega L || R)$$, where $$||$$ represents "parallel." (i.e. $$j \omega L$$ in parallel with $$R$$.)

$$Z_{EQ} = R + \frac{j \omega LR}{j \omega L + R}$$

I stopped there because I realized I would be nowhere near close to the correct answer. Am I even on the right track with this logic first of all?

$$I_{L}= \frac{V_{A}}{R+2j \omega L}$$
$$V_{L}=\frac{j \omega L V_{A}}{R+2j \omega L}$$

Joined Feb 16, 2010
112
You were headed in the right direction with that idea.

Converting to the frequency domain, you get:
$$Z=R+(jwL||R)=R+jwLR/(jwL+R)=\frac{R(jwL+R)+jwLR}{jwL+R} Itotal=\frac{Va}{Z}=\frac{Va(jwL+R)}{R^{2}+2jwLR}$$

Next you use a current divider between jwL and R:
$$I=Itotal*\frac{R}{jwL+R}=\frac{Va(jwL+R)}{R^{2}+2jwLR}*\frac{R}{jwL+R}=\frac{Va}{R+2jwL}$$

Try the voltage yourself and don't give up prematurely because you think you might be wrong.